Problem Description
Clarke is a patient with multiple personality disorder. One day, Clarke split into two personality a and b, they are playing a game. 
There is a n∗m matrix, each grid of this matrix has a number ci,j. 
a wants to beat b every time, so a ask you for a help. 
There are q operations, each of them is belonging to one of the following two types: 
1. They play the game on a (x1,y1)−(x2,y2) sub matrix. They take turns operating. On any turn, the player can choose a grid which has a positive integer from the sub matrix and decrease it by a positive integer which less than or equal this grid's number. The player who can't operate is loser. a always operate first, he wants to know if he can win this game. 
2. Change ci,j to b.

 
Input
The first line contains a integer T(1≤T≤5), the number of test cases. 
For each test case: 
The first line contains three integers n,m,q(1≤n,m≤500,1≤q≤2∗105) 
Then n∗m matrix follow, the i row j column is a integer ci,j(0≤ci,j≤109) 
Then q lines follow, the first number is opt. 
if opt=1, then 4 integers x1,y1,x1,y2(1≤x1≤x2≤n,1≤y1≤y2≤m) follow, represent operation 1. 
if opt=2, then 3 integers i,j,b follow, represent operation 2.

 

 
Output
For each testcase, for each operation 1, print Yes if a can win this game, otherwise print No.

 

Sample Input
1
1 2 3
1 2
1 1 1 1 2
2 1 2 1
1 1 1 1 2

 

Sample Output
Yes 
No

 

Hint: The first enquiry: $a$ can decrease grid $(1, 2)$'s number by $1$. No matter what $b$ operate next, there is always one grid with number $1$ remaining . So, $a$ wins. The second enquiry: No matter what $a$ operate, there is always one grid with number $1$ remaining. So, $b$ wins.
 

 

Source
 

 题目要求二维的nim游戏，考虑到nim的结论是xor和为0则必败、否则必胜，那么我们只需要维护子矩阵的xor和。由于xor有前缀和性质，所以我们可以用一个二维bit来维护(1, 1)-(a, b)的矩阵的xor和，然后由sum(x2,y2) xor sum(x2,y1−1) xor sum(x1−1,y2) xor sum(x1−1,y1−1)sum(x2, y2) \ xor \ sum(x2, y1-1) \ xor \ sum(x1-1, y2) \ xor \ sum(x1-1, y1-1)sum(x2,y2) xor sum(x2,y1−1) xor sum(x1−1,y2) xor sum(x1−1,y1−1)来得到答案即可。单点修改在bit上是很容易的。

 

hdu 5465 Clarke and puzzle(前缀和，异或，nim博弈)_#includehdu 5465 Clarke and puzzle(前缀和，异或，nim博弈)_#include_02
 1 #pragma comment(linker, "/STACK:1024000000,1024000000")
 2 #include<iostream>
 3 #include<cstdio>
 4 #include<cstring>
 5 #include<cmath>
 6 #include<math.h>
 7 #include<algorithm>
 8 #include<queue>
 9 #include<set>
10 #include<bitset>
11 #include<map>
12 #include<vector>
13 #include<stdlib.h>
14 #include <stack>
15 using namespace std;
16 int dirx[]={0,0,-1,1};
17 int diry[]={-1,1,0,0};
18 #define PI acos(-1.0)
19 #define max(a,b) (a) > (b) ? (a) : (b)  
20 #define min(a,b) (a) < (b) ? (a) : (b)
21 #define ll long long
22 #define eps 1e-10
23 #define MOD 1000000007
24 #define N 506
25 #define inf 1e12
26 int n,m,q;
27 int mp[N][N];
28 int a[N][N];
29 int main()
30 {
31     int t;
32     scanf("%d",&t);
33     while(t--){
34         scanf("%d%d%d",&n,&m,&q);
35         for(int i=1;i<=n;i++){
36             for(int j=1;j<=m;j++){
37                 scanf("%d",&mp[i][j]);
38                 a[i][j]=a[i][j-1]^mp[i][j];
39             }
40         }
41         for(int i=0;i<q;i++){
42             int opt;
43             scanf("%d",&opt);
44             int x,y,z;
45             if(opt==2){
46                 scanf("%d%d%d",&x,&y,&z);
47                 mp[x][y]=z;
48                 for(int j=y;j<=m;j++){
49                     a[x][j]=a[x][j-1]^mp[x][j];
50                 }
51             }
52             else{
53                 int ans=0;
54                 int x1,y1,x2,y2;
55                 scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
56                 for(int j=x1;j<=x2;j++){
57                     ans=ans^(a[j][y2]^a[j][y1-1]);
58                 }
59                 if(ans==0){
60                     printf("No\n");
61                 }
62                 else{
63                     printf("Yes\n");
64                 }
65             }
66         }
67     }
68     return 0;
69 }
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